To cool 250 mL of coffee at 90.0°C, you put a 80 g metal spoon cooled to 0.0°C in the coffee. After thermal equilibrium has been reached, what is the final temperature of your coffee? Assume energy is exchanged only between the spoon and the coffee. The specific heat capacity of the metal spoon is 0.80 J/(g°C). Assume the specific heat and density of coffee are the same as those of water, 4.18 J/(g · °C) and 1.00 g/mL. Keep one decimal in your answer.

The correct answer is 84.81 °C.

Explanation:

Based on the given information, the volume or mass of coffee is 250 ml, and the mass of metal spoon is 80 grams. The specific heat capacity of metal spoon is 0.80 J/g °C, and the specific heat capacity of coffee is 4.18 J/g °C. In the given case, heat is lost from the coffee and is gained by the spoon.  

For finding the final temperature of the coffee, the formula to be used is Q = msΔT, here m is the mass, s is the specific heat, and ΔT is the change in temperature.  

Heat lost from the coffee = Heat gained by the spoon

250 × 4.18 (90 - T2) = 80 × 0.8 (T2 - 0)

1045 (90 - T2) = 64 (T2 -0)

94050 - 1045 T2 = 64 T2

94050 = 1045 T2 + 64 T2

94050 = 1109 T2

T2 = 84.84 °C

a gas, having a molecular weight of 13 kg/kmol and a kinematic viscosity of 0.25 cm2/s, flows through a pipe 0.25 m internal diameter and 5 km long at the rate of 0.4 m3/s and is delivered at atmospheric pressure.